Question: Simplify and expand the following expression: $ \dfrac{x + 6}{2x + 1}-\dfrac{4x}{2x - 1} $
Answer: In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(2x + 1)(2x - 1)$ Multiply the first term by $\dfrac{2x - 1}{2x - 1}$ $ \begin{align*} \dfrac{x + 6}{2x + 1} \times \dfrac{2x - 1}{2x - 1} & = \dfrac{(x + 6)(2x - 1)}{(2x + 1)(2x - 1)} \\ & = \dfrac{2x^2 + 11x - 6}{(2x + 1)(2x - 1)}\end{align*} $ Multiply the second term by $\dfrac{2x + 1}{2x + 1}$ $ \begin{align*} \dfrac{4x}{2x - 1} \times \dfrac{2x + 1}{2x + 1} & = \dfrac{(4x)(2x + 1)}{(2x - 1)(2x + 1)} \\ & = \dfrac{8x^2 + 4x}{(2x - 1)(2x + 1)}\end{align*} $ Now we have: $ = \dfrac{2x^2 + 11x - 6}{(2x + 1)(2x - 1)} - \dfrac{8x^2 + 4x}{(2x - 1)(2x + 1)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{2x^2 + 11x - 6 - (8x^2 + 4x)}{(2x + 1)(2x - 1)} $ $ = \dfrac{2x^2 + 11x - 6 - 8x^2 - 4x}{(2x + 1)(2x - 1)} $ $ = \dfrac{-6x^2 + 7x - 6}{(2x + 1)(2x - 1)}$ Expand the denominator: $ = \dfrac{-6x^2 + 7x - 6}{4x^2 - 1}$